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[Leetcode]52. N-Queens II(C++)

题目描述

题目链接:52. N-Queens II

The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.

Given an integer n, return the number of distinct solutions to the n-queens puzzle.

例子

例子 1

见官网例子。

Constraints

  • 1 <= n <= 9

解题思路

思路和 [Leetcode]51. N-Queens(C++) 基本一样,只是不需要存储具体的解,只需要用一个整型存储解的数量,在找到解时将其加一即可。代码如下:

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#include <string>
#include <unordered_set>
#include <vector>

class Solution {
public:
int totalNQueens(int n) {
int result = 0;
std::unordered_set<int> col_used;
std::unordered_set<int> diag1;
std::unordered_set<int> diag2;
dfs(0, n, col_used, diag1, diag2, result);
return result;
}

private:
void dfs(int row, int n, std::unordered_set<int>& col_used,
std::unordered_set<int>& diag1, std::unordered_set<int>& diag2,
int& result) {
if (row == n) {
result++;
return;
}

std::string curr_row(n, '.');
for (int col = 0; col < n; ++col) {
if (col_used.count(col) || diag1.count(col - row) ||
diag2.count(col + row))
continue;
curr_row[col] = 'Q';
col_used.insert(col);
diag1.insert(col - row);
diag2.insert(col + row);
dfs(row + 1, n, col_used, diag1, diag2, result);
col_used.erase(col);
diag1.erase(col - row);
diag2.erase(col + row);
}
}
};
  • 时间复杂度: O(n!)
  • 空间复杂度: O(n)

GitHub 代码同步地址: 52.NQueensIi.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions